package com.hit.basmath.learn.others;

/**
 * 419. Battleships in a Board
 * <p>
 * Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules:
 * <p>
 * 1. You receive a valid board, made of only battleships or empty slots.
 * 2. Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.
 * 3. At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
 * <p>
 * Example:
 * <p>
 * X..X
 * ...X
 * ...X
 * <p>
 * In the above board there are 2 battleships.
 * <p>
 * Invalid Example:
 * <p>
 * ...X
 * XXXX
 * ...X
 * <p>
 * This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
 * <p>
 * Follow up:
 * <p>
 * Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?
 */
public class _419 {
    public int countBattleships(char[][] board) {
        int count = 0;
        for (int i = 0; i < board.length; i++)
            for (int j = 0; j < board[0].length; j++)
                if (board[i][j] == 'X') {
                    boolean k = true;
                    if (i != 0)
                        if (board[i - 1][j] == 'X')
                            k = false;
                    if (j != 0)
                        if (board[i][j - 1] == 'X')
                            k = false;
                    if (k == true)
                        count++;
                }
        return count;
    }
}
